Elements of Geometry and Trigonometry |
From inside the book
Results 6-10 of 38
Page 40
... corresponding terms will be proportional . Let and then will For since and or therefore , M : NP : Q RST : V : MXR NXS :: PxT : QxV . MxQ = NxP RXV SXT , we shall have MxQxRxV = NxPxSxT MXRxQxV = NxSxPxT MXR NXS :: P × T : Q × V . BOOK ...
... corresponding terms will be proportional . Let and then will For since and or therefore , M : NP : Q RST : V : MXR NXS :: PxT : QxV . MxQ = NxP RXV SXT , we shall have MxQxRxV = NxPxSxT MXRxQxV = NxSxPxT MXR NXS :: P × T : Q × V . BOOK ...
Page 51
... corresponding arc to the corresponding arc . Suppose , for example , that the angles ACB , BOOK III . 51.
... corresponding arc to the corresponding arc . Suppose , for example , that the angles ACB , BOOK III . 51.
Page 68
... correspond to equal angles at the centre . Thus , if the angles A and O are equal , the arc BC will be similar to DE , the sector BAC to the sector DOE , and the segment whose chord is BC , to the seg- ment whose chord is DE . ДД 4. The ...
... correspond to equal angles at the centre . Thus , if the angles A and O are equal , the arc BC will be similar to DE , the sector BAC to the sector DOE , and the segment whose chord is BC , to the seg- ment whose chord is DE . ДД 4. The ...
Page 72
... corresponding terms of these proportions together , and observing that the term AEHD may be omit- ted , since it is a multiplier of both the antecedent and the con- sequent , we shall have ABCD AEGF :: ABX AD : AEX AF . Scholium . Hence ...
... corresponding terms of these proportions together , and observing that the term AEHD may be omit- ted , since it is a multiplier of both the antecedent and the con- sequent , we shall have ABCD AEGF :: ABX AD : AEX AF . Scholium . Hence ...
Page 81
... corresponding members together , and observing that BE and DE are equal , we shall have AB2 + AD2 + DC2 + BC2¬4AE2 + 4DE2 . But 4AE2 is the square of 2AE , or of AC ; 4DE2 is the square of BD ( Prop . VIII . Cor . ) ; hence the squares ...
... corresponding members together , and observing that BE and DE are equal , we shall have AB2 + AD2 + DC2 + BC2¬4AE2 + 4DE2 . But 4AE2 is the square of 2AE , or of AC ; 4DE2 is the square of BD ( Prop . VIII . Cor . ) ; hence the squares ...
Other editions - View all
Common terms and phrases
adjacent adjacent angles altitude angle ACB angle BAC ar.-comp base multiplied bisect Book centre chord circ circumference circumscribed common cone consequently convex surface cylinder diagonal diameter dicular distance draw drawn equal angles equally distant equation equiangular equivalent figure formed four right angles frustum given angle given line gles greater homologous sides hypothenuse inscribed circle inscribed polygon intersection less Let ABC let fall logarithm measured by half number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedron polygon ABCDE prism proportional PROPOSITION pyramid quadrant quadrilateral quantities radii radius ratio rectangle regular polygon right angled triangle S-ABC Scholium secant secant line segment side BC similar solid angle solid described sphere spherical polygon spherical triangle square described straight line tang tangent THEOREM triangle ABC triangular prism vertex